An Easy Integration,Is It?

Calculus Level 5

Evaluate

∫ − 1 3 1 3 x 4 1 − x 4 ⋅ cos ⁡ − 1 ( 2 x 1 + x 2 ) d x \displaystyle \large \int^{\frac{1}{\sqrt{3}}}_{\frac{-1}{\sqrt{3}}} \dfrac{x^4}{1-x^4}\cdot \cos^{-1}(\dfrac{2x}{1+x^2}) \mathrm{dx}

If it can be expressed as

π a ln ⁡ ( b + c ) + π d e − π f \displaystyle \dfrac{\pi}{a}\ln(b+\sqrt{c}) +\dfrac{\pi^{d}}{e} - \dfrac{\pi}{\sqrt{f}}

Then Find

abcdef + 1 \large \text{abcdef} + 1


The answer is 1729.

Vraj Mehta by Vraj Mehta , 30, India

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2 solutions

This problem is brilliantly engineered to give answer as 1729 1729 . It's pretty simple too.

I = ∫ − 1 3 1 3 x 4 1 − x 4 cos ⁡ − 1 2 x 1 + x 2 d x p u t − x = t , ⇒ I = ∫ − 1 3 1 3 t 4 1 − t 4 ( π − cos ⁡ − 1 2 t 1 + t 2 ) d t ⇒ 2 I = π ∫ − 1 3 1 3 x 4 1 − x 4 d x I = π 2 ∫ − 1 3 1 3 x 4 − 1 + 1 1 − x 4 d x = π 2 ∫ − 1 3 1 3 − 1 + 1 1 − x 4 d x = π 2 ∫ − 1 3 1 3 − 1 + 1 − x 2 + 1 + x 2 2 ( 1 − x 2 ) ( 1 + x 2 ) d x = π 2 ∫ − 1 3 1 3 − 1 + 1 2 ( 1 − x 2 ) + 1 2 ( 1 + x 2 ) d x = π 2 [ − 2 3 + 1 2 ln ⁡ ∣ 1 + 3 1 − 3 ∣ + π 6 ] I = π 4 ln ⁡ ( 2 + 3 ) + π 2 12 − π 3 ⇒ a b c d e f + 1 = 1728 + 1 = 1729 \displaystyle \large I=\int _{ \frac { -1 }{ \sqrt { 3 } } }^{ \frac { 1 }{ \sqrt { 3 } } }{ \frac { { x }^{ 4 } }{ 1-{ x }^{ 4 } } \cos ^{ -1 }{ \frac { 2x }{ 1+{ x }^{ 2 } } } } dx\\ put\quad -x=t,\quad \\ \Rightarrow I=\int _{ \frac { -1 }{ \sqrt { 3 } } }^{ \frac { 1 }{ \sqrt { 3 } } }{ \frac { { t }^{ 4 } }{ 1-{ t }^{ 4 } } (\pi -\cos ^{ -1 }{ \frac { 2t }{ 1+{ t }^{ 2 } } } ) } dt\\ \Rightarrow 2I=\pi \int _{ \frac { -1 }{ \sqrt { 3 } } }^{ \frac { 1 }{ \sqrt { 3 } } }{ \frac { { x }^{ 4 } }{ 1-{ x }^{ 4 } } } dx\\ I=\frac { \pi }{ 2 } \int _{ \frac { -1 }{ \sqrt { 3 } } }^{ \frac { 1 }{ \sqrt { 3 } } }{ \frac { { x }^{ 4 }-1+1 }{ 1-{ x }^{ 4 } } } dx=\frac { \pi }{ 2 } \int _{ \frac { -1 }{ \sqrt { 3 } } }^{ \frac { 1 }{ \sqrt { 3 } } }{ -1+\frac { 1 }{ 1-{ x }^{ 4 } } } dx\\ =\frac { \pi }{ 2 } \int _{ \frac { -1 }{ \sqrt { 3 } } }^{ \frac { 1 }{ \sqrt { 3 } } }{ -1+\frac { 1-{ x }^{ 2 }+1+{ x }^{ 2 } }{ 2(1-{ x }^{ 2 })(1+{ x }^{ 2 }) } } dx\\ =\frac { \pi }{ 2 } \int _{ \frac { -1 }{ \sqrt { 3 } } }^{ \frac { 1 }{ \sqrt { 3 } } }{ -1+\frac { 1 }{ 2(1-{ x }^{ 2 }) } +\frac { 1 }{ 2(1+{ x }^{ 2 }) } } dx\\ =\frac { \pi }{ 2 } \left[ \frac { -2 }{ \sqrt { 3 } } +\frac { 1 }{ 2 } \ln { \left| \frac { 1+\sqrt { 3 } }{ 1-\sqrt { 3 } } \right| } +\frac { \pi }{ 6 } \right] \\ I=\frac { \pi }{ 4 } \ln { (2+\sqrt { 3 } ) } +\frac { { \pi }^{ 2 } }{ 12 } -\frac { \pi }{ \sqrt { 3 } } \\ \Rightarrow abcdef+1=1728+1=\boxed { 1729 } \\ \\

@Vraj Mehta

14 Helpful 0 Interesting 1 Brilliant 0 Confused

Yeah Bringing That answer Took More Time Than Solving

Vraj Mehta - 6 years, 3 months ago

but i did the same yet it is 4 2 3 12 3 = 864 ? ? ans = 865 ??

A Former Brilliant Member - 4 years, 4 months ago

Nice problem with a nice answer ! : ) :)

Keshav Tiwari - 6 years ago
Bhargav Upadhyay
Mar 4, 2015

I = ∫ − 1 3 1 3 ( x 4 ( x 4 − 1 ) ) ( cos ⁡ − 1 ( 2 x 1 + x 2 ) ) d x = ∫ − 1 3 1 3 ( x 4 ( x 4 − 1 ) ) ( π 2 − sin ⁡ − 1 ( 2 x 1 + x 2 ) ) d x I = π 2 ∫ − 1 3 1 3 ( x 4 ( x 4 − 1 ) ) d x { ∵ sin ⁡ − 1 x i s o d d f u n c t i o n } I=\int _{ \frac { -1 }{ \sqrt { 3 } } }^{ \frac { 1 }{ \sqrt { 3 } } }{ (\frac { { x }^{ 4 } }{ ({ x }^{ 4 }-1) } )\quad (\cos ^{ -1 }{ (\frac { 2x }{ 1+{ x }^{ 2 } } ) } ) } dx\\ =\int _{ \frac { -1 }{ \sqrt { 3 } } }^{ \frac { 1 }{ \sqrt { 3 } } }{ (\frac { { x }^{ 4 } }{ ({ x }^{ 4 }-1) } )\quad (\frac { \pi }{ 2 } -\sin ^{ -1 }{ (\frac { 2x }{ 1+{ x }^{ 2 } } ) } ) } dx\\ I=\frac { \pi }{ 2 } \int _{ \frac { -1 }{ \sqrt { 3 } } }^{ \frac { 1 }{ \sqrt { 3 } } }{ (\frac { { x }^{ 4 } }{ ({ x }^{ 4 }-1) } ) } dx\quad \{ \because \quad \sin ^{ -1 }{ x } \quad is\quad odd\quad function\} \\

Now same as Raghav Vaidyanathan solution

2 Helpful 0 Interesting 0 Brilliant 0 Confused

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